Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{x^2 - 3x - 54}{-2x^2 + 4x + 160} \div \dfrac{x + 6}{4x - 40} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{x^2 - 3x - 54}{-2x^2 + 4x + 160} \times \dfrac{4x - 40}{x + 6} $ First factor out any common factors. $q = \dfrac{x^2 - 3x - 54}{-2(x^2 - 2x - 80)} \times \dfrac{4(x - 10)}{x + 6} $ Then factor the quadratic expressions. $q = \dfrac {(x + 6)(x - 9)} {-2(x - 10)(x + 8)} \times \dfrac {4(x - 10)} {x + 6} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (x + 6)(x - 9) \times 4(x - 10)} { -2(x - 10)(x + 8) \times (x + 6)} $ $q = \dfrac {4(x + 6)(x - 9)(x - 10)} {-2(x - 10)(x + 8)(x + 6)} $ Notice that $(x - 10)$ and $(x + 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {4(x + 6)(x - 9)\cancel{(x - 10)}} {-2\cancel{(x - 10)}(x + 8)(x + 6)} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $q = \dfrac {4\cancel{(x + 6)}(x - 9)\cancel{(x - 10)}} {-2\cancel{(x - 10)}(x + 8)\cancel{(x + 6)}} $ We are dividing by $x + 6$ , so $x + 6 \neq 0$ Therefore, $x \neq -6$ $q = \dfrac {4(x - 9)} {-2(x + 8)} $ $ q = \dfrac{-2(x - 9)}{x + 8}; x \neq 10; x \neq -6 $